Question 1018487
You can find the "altitude" of one of the slant triangles, and call this L.  The slant triangle is an isosceles triangle.


{{{(3/2)^2+L^2=5^2}}}
{{{L^2=5^2-(3/2)^2}}}
{{{L=sqrt(5^2-(3/2)^2)}}}



The altitude of the pyramid is what you want next.  A right triangle crossing through the pyramid altitude and a slant-altitude.  Pythagorean Theorem formula again.  
According to the figure,
{{{x^2+(3/2)^2=L^2}}}
{{{x^2=L^2-(3/2)^2}}}
{{{highlight(x=sqrt(L^2-(3/2)^2))}}}


You can substitute there for L^2 as found earlier, although was not simplified there;
{{{highlight(x=sqrt(5^2-(3/2)^2-(3/2)^2))}}}, and THIS you can now simplify and compute.